在新列中赋值——请提高效率

Question

我正在寻找一种更有效的方法来完成一项非常基本的任务：添加一个新列，其中包含为现有行指定的值。示例数据框（称为 ess）具有国家和（调查）回合。我想添加一列“dem”，其中包含来自外部来源的值。这是一个片段：

id cntry essround dem
1 AL       1
2 AT       1
3 BE       1
4 BG       1
5 HR       1
6 AL       2
7 AT       2
8 BE       2
9 BG       2
10 HR      2

执行此操作的“长”方法如下：

ess$dem <- NA
ess$dem[ess$cntry=="AL" & ess$essround==1] <- 3.5
ess$dem[ess$cntry=="AT" & ess$essround==1] <- 1
ess$dem[ess$cntry=="BE" & ess$essround==1] <- 1.5
ess$dem[ess$cntry=="BG" & ess$essround==1] <- 2
ess$dem[ess$cntry=="HR" & ess$essround==1] <- 2
ess$dem[ess$cntry=="AL" & ess$essround==2] <- 3
ess$dem[ess$cntry=="AT" & ess$essround==2] <- 1
ess$dem[ess$cntry=="BE" & ess$essround==2] <- 1
ess$dem[ess$cntry=="BG" & ess$essround==2] <- 1.5
ess$dem[ess$cntry=="HR" & ess$essround==2] <- 2

问题是，当我有 36 个国家和 6 个回合时，这种方法变得非常长——这样我最终得到 216 行代码。（当我想在同一模式下创建多个新列时，情况会变得更糟...）

有没有办法压缩这样的操作？？是否可以在代码依赖于相应值列表中的“位置”的单行上完成？

Answer 1

正在创建虚拟数据：

ess = data.frame(
contry = sample(c("AL","AT","BE","BG","HR","AL","AT","BE","BG","HR"), 20, TRUE),
essround = sample(1:2, 20, TRUE))

现在代码：

ess$dem <- NA

values = c(3.5,1,1.5,2,2,3,1,1,1.5,2)
groups = unique(ess$contry)

for(i in 1:length(groups)){
  ess[ess$contry==groups[i],"dem"] <- values[i]
}

输出：

   contry essround dem
1      BE        1 3.5
2      HR        2 1.0
3      AT        2 1.5
4      BG        1 2.0
5      AT        1 1.5
6      AT        2 1.5
7      AT        2 1.5
8      AT        2 1.5
9      BG        2 2.0
10     BE        2 3.5
11     AT        1 1.5
12     AT        2 1.5
13     BE        1 3.5
14     AT        2 1.5
15     HR        1 1.0
16     BG        1 2.0
17     BE        1 3.5
18     BG        1 2.0
19     AT        2 1.5
20     AT        2 1.5

Answer 2

首先使用 tidyverse，您需要创建一个包含值的 data.frame：即 ess$dem[ess$cntry=="AL" & ess$essround==1] <- 3.5 应该成为 conds data.frame 中的一行:

## expand grid to create all possible combinations of cntry and essround
conds <- expand.grid(cntry=c("AL","AT","BE","BG","HR"), essround=1:2) %>% mutate(dem = c(3.5,1,1.5,2,2,3,1,1,1.5,2))
## first row will be "AL" 1 3.5 which is the first condition
conds
   cntry essround dem
1     AL        1 3.5
2     AT        1 1.0
3     BE        1 1.5
4     BG        1 2.0
5     HR        1 2.0
6     AL        2 3.0
7     AT        2 1.0
8     BE        2 1.0
9     BG        2 1.5
10    HR        2 2.0
ess %>% left_join(conds)
Joining, by = c("cntry", "essround")
   cntry essround dem
1     AT        1 1.0
2     AT        2 1.0
3     HR        2 2.0
4     BG        2 1.5
5     HR        2 2.0
6     HR        1 2.0
7     BG        2 1.5
8     BG        1 2.0
9     HR        2 2.0
10    BG        1 2.0
11    AT        1 1.0
12    BG        2 1.5
13    AL        1 3.5
14    HR        1 2.0
15    BE        2 1.0
16    AL        2 3.0
17    AL        1 3.5
18    AL        1 3.5
19    AT        1 1.0
20    AT        1 1.0

在新列中赋值——请提高效率

assigning values in a new column -- efficiency, please

r

dataframe

assign